Calculating $\int_0^{\pi/4} (2\cos^2 x-1)^{3/2} \sin x \,dx$ I'm having a hard time trying to calculate the integral
$$
\int_0^{\pi/4} (2\cos^2 x-1)^{3/2} \sin x \,dx.
$$
I have tried the following: $z=
\cos x$,
$$
\int_0^{\pi/4} (2\cos^2 x-1)^{3/2} \sin x \,dx= -\int_1^{\sqrt2/2} (2z^2-1)^{3/2} \,dz
$$
introduce the new variable $z=1/\sqrt2\cos t$, $dz=-\frac1{\sqrt2}\cos^{-2} t(-\sin t)\, dt$:
$$
=-\frac1{\sqrt2}\int_{\pi/4}^{0} \tan^3 t  \cos^{-2} t\sin t\, dt
=-\frac1{\sqrt2}\int_{\pi/4}^{0} \frac{\sin^4 t}{\cos^5 t}\, dt
$$
$$
=-\frac1{\sqrt2}\int_{\pi/4}^{0} \frac{\sin^4 t}{\cos^6 t}\cos t\, dt
=-\frac1{\sqrt2}\int_{\pi/4}^{0} \frac{\sin^4 t}{(1-\sin^2 t)^3}\, d(\sin t)
$$
(the next variable is $u=\sin t$)
$$
=-\frac1{\sqrt2}\int_{\pi/4}^{0} \frac{u^4}{(1-u^2)^3}\, du
$$
This integral scared me so much that I began to doubt whether I was on the right way. Should I use the method of undetermined coefficients or am I missing something?
 A: A slightly less scary sub in the second step is $2u^2 = \cosh^2 y$. Note that $\cosh^2 y - 1 =\sinh^2 y$ and $d(\cosh y) = \sinh y \ dy$.
The hyperbolic substitution allows you to recast the integral as a constant times $\sinh^4 y$.
By writing $\displaystyle \sinh y = \frac{e^y - e^{-y}}2$ you can use binomial theorem to write $\sinh^4 y$ in terms of exponentials which are easy to integrate.
A: In order to compute$$\int(2z^2-1)^{3/2}\,\mathrm dz\tag1,$$you can put $z=\frac{\sec t}{\sqrt2}$ and $\mathrm dz=\frac1{\sqrt2}\sec(t)\tan(t)\,\mathrm dt$. So, $(1)$ becomes\begin{align}\frac1{\sqrt2}\int\tan^4(t)\sec(t)\,\mathrm dt&=\frac1{\sqrt2}\int\frac{\sin^4t}{\cos^5t}\,\mathrm dt\\&=\frac1{\sqrt2}\int\frac{\sin^4t}{(1-\sin^2t)^3}\cos t\,\mathrm dt.\end{align}This leads you to the computation of a primitive of $\frac{x^4}{(1-x^2)^3}$; you can take$$\frac1{16}\left(\frac{2x\left(5 x^2-3\right)}{\left(x^2-1\right)^2}-3\log (1-x)+3\log (x+1)\right).$$
A: Let $\cosh t= \sqrt2\cos x$. Then,
$$\begin{align}
& \int_0^{\pi/4} (2\cos^2 x-1)^{3/2} \sin x \,dx \\
&=\frac1{\sqrt2}\int^{\cosh^{-1}1}_{\cosh^{-1}\sqrt2} \sinh^4 tdt\\
&=\frac1{32\sqrt2}(12t-8\sinh2t+\sinh4t)\bigg|^{\cosh^{-1}1}_{\cosh^{-1}\sqrt2}\\
&=\frac1{16}(2-3\sqrt2\cosh^{-1}\sqrt2)\\
\end{align}$$
A: Let $\sqrt2=a$
We need
$$a\int(z^2-a^2)^{3/2}dz$$
Using $\#15$ of this and setting $n=-\dfrac12\iff2n=-1$
$$\int\dfrac{dz}{\sqrt{z^2-a^2}}=-\dfrac z{2(-1/2-1)a^2(z^2-a^2)^{-1/2-1}}-\dfrac{-1-3}{(-1-2)a^2}\int\dfrac{dz}{(z^2-a^2)^{-1/2-1}}$$
$$\implies\int\dfrac{dz}{(z^2-a^2)^{-1/2-1}}=?$$
Again $$\int\dfrac{dz}{\sqrt{z^2-a^2}}=\ln(z+\sqrt{z^2-a^2})+K$$
